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Course by: Steven J. Cox. E-mail the author

# The Null and Column Spaces: An Example

Module by: Doug Daniels, Steven J. Cox. E-mail the authors

Summary: In this module we compute bases for the null and column spaces of an adjacency matrix associated with a ladder.

## Preliminary Information

Let us compute bases for the null and column spaces of the adjacency matrix associated with the ladder below.

The ladder has 8 bars and 4 nodes, so 8 degrees of freedom. Denoting the horizontal and vertical displacements of node jj by x 2 j 1 x 2 j 1 and x 2 j x 2 j , respectively, we arrive at the AA matrix A=( 10000000 -10100000 00-100000 0-1000100 000-10001 00001000 0000-1010 000000-10 ) A 1000 0000 -1010 0000 00-10 0000 0-100 0100 000-1 0001 0000 1000 0000 -1010 0000 00-10

## Finding a Basis for the Column Space

To determine a basis for A A we must find a way to discard its dependent columns. A moment's reflection reveals that columns 2 and 6 are colinear, as are columns 4 and 8. We seek, of course, a more systematic means of uncovering these and perhaps other less obvious dependencies. Such dependencies are more easily discerned from the row reduced form A red =rrefA=( 10000000 01000-100 00100000 0001000-1 00001000 00000010 00000000 00000000 ) A red rref A 1000 0000 0100 0-100 0010 0000 0001 000-1 0000 1000 0000 0010 0000 0000 0000 0000 Recall that rref performs the elementary row operations necessary to eliminate all nonzeros below the diagonal. For those who can't stand to miss any of the action I recommend rrefmovie.

Each nonzero row of A red A red is called a pivot row. The first nonzero in each row of A red A red is called a pivot. Each column that contains a pivot is called a pivot column. On account of the staircase nature of A red A red we find that there are as many pivot columns as there are pivot rows. In our example there are six of each and, again on account of the staircase nature, the pivot columns are the linearly independent columns of A red A red . One now asks how this might help us distinguish the independent columns of AA. For, although the rows of A red A red are linear combinations of the rows of AA, no such thing is true with respect to the columns. In our example, columns 123457 1 2 3 4 5 7 are the pivot columns. In general:

### Proposition 1

Suppose AA is m-by-n. If columns c j j= 1 , ... , r c j j 1 , ... , r are the pivot columns of A red A red . then columns c j j= 1 , ... , r c j j 1 , ... , r of AA constitute a basis for A A .

#### Proof

Note that the pivot columns of A red A red . are, by construction, linearly independent. Suppose, however, that columns c j j= 1 , ... , r c j j 1 , ... , r of AA are linearly dependent. In this case there exists a nonzero x n x n for which Ax=0 A x 0 and

x k =0  ,   k c j j= 1 , ... , r    k k c j j 1 , ... , r x k 0
(1)
Now Ax=0 A x 0 necessarily implies that A red x=0 A red x 0 , contrary to the fact that columns c j j= 1 , ... , r c j j 1 , ... , r are the pivot columns of A red A red .

We now show that the span of columns c j j= 1 , ... , r c j j 1 , ... , r of AA indeed coincides with A A . This is obvious if r=n r n , i.e., if all of the columns are linearly independent. If r<n r n , there exists a q c j j= 1 , ... , r q c j j 1 , ... , r . Looking back at A red A red . we note that its qqth column is a linear combination of the pivot columns with indices not exceeding qq. Hence, there exists an xx satisfying Equation 1 and A red x=0 A red x 0 , and x q =1 x q 1 . This xx then necessarily satisfies Ax=0 A x 0 . This states that the qqth column of AA is a linear combination of columns c j j= 1 , ... , r c j j 1 , ... , r of AA.

## Finding a Basis for the Null Space

Let us now exhibit a basis for 𝒩A 𝒩 A . We exploit the already mentioned fact that 𝒩A=𝒩 A red 𝒩 A 𝒩 A red . Regarding the latter, we partition the elements of xx into so called pivot variables, x c j j= 1 , ... , r x c j j 1 , ... , r and free variables x k k c j j= 1 , ... , r x k k c j j 1 , ... , r There are evidently nr n r free variables. For convenience, let us denote these in the future by x c j j= r + 1 , ... , n x c j j r + 1 , ... , n

One solves A red x=0 A red x 0 , by expressing each of the pivot variables in terms of the nonpivot, or free, variables. In the example above, x 1 x 1 , x 2 x 2 , x 3 x 3 , x 4 x 4 , x 5 x 5 , and x 7 x 7 are pivot while x 6 x 6 and x 8 x 8 are free. Solving for the pivot in terms of the free we find x 7 =0 x 7 0 x 5 =0 x 5 0 x 4 = x 8 x 4 x 8 x 3 =0 x 3 0 x 2 = x 6 x 2 x 6 x 1 =0 x 1 0 or, written as a vector,

x= x 6 01000100+ x 8 00010001 x x 6 0100 0100 x 8 0001 0001
(2)
where x 6 x 6 and x 8 x 8 are free. As x 6 x 6 and x 8 x 8 range over all real numbers, the xx above traces out a plane in 8 8 . This plane is precisely the null space of AA and Equation 2 describes a generic element as the linear combination of two basis vectors. Compare this to what MATLAB returns when faced with null(A,'r'). Abstracting these calculations we arrive at

### Proposition 2

Suppose that AA is m-by-n with pivot indices c j j= 1 , ... , r c j j 1 , ... , r and free indices c j j= r + 1 , ... , n c j j r + 1 , ... , n . A basis for 𝒩A 𝒩 A . may be constructed of nr n r vectors z 1 z 2 ... z n - r z 1 z 2 ... z n - r where z k z k , and only z k z k , possesses a nonzero in its c r + k c r + k component.

## The Physical Meaning of Our Calculations

Let us not end on an abstract note however. We ask what A A and 𝒩A 𝒩 A tell us about the ladder. Regarding A A the answer will come in the next chapter. The null space calculation however has revealed two independent motions against which the ladder does no work! Do you see that the two vectors in Equation 2 encode rigid vertical motions of bars 4 and 5 respectively? As each of these lies in the null space of AA, the associated elongation is zero. Can you square this with the ladder as pictured in Figure 1? I hope not, for vertical motion of bar 4 must 'stretch' bars 1, 2, 6, and 7. How does one resolve this (apparent) contradiction?

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